Two blocks are connected over a massless pulley as shown in figure. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2 Block A sliders down the incline at constant speed. The mass of block B in kg is

Options

A. 5.4
B. 3.3
C. 4.2
D. 6.8

Answer

B

Explanation

(b) Net pulling force on the system should be zero, as velocity is constant. Hence,
`m_(A)g sin 30^(@)=mu m_(A)g cos 30^(@)+m_(B)g`
`therefore" " m_(B)=((m_(A))/(2))-((mu m_(A)sqrt(3))/(2))`
`=10[(1)/(2)-0.2xx(sqrt(3))/(2)]=3.3 kg `